# Proving the slopes to the left and right of a maximum

Consider a continuous and differentiable function $f$ with a maximum point at $x=c$. I want to show that

$$\lim_{h \to 0-}\frac{f(c+h) - f(c)}{h} \geq 0$$

$$\lim_{h \to 0+}\frac{f(c+h) - f(c)}{h} \leq 0$$

must be true. But when I tried to set up the typical epsilon-delta formulation I very quickly got stuck.

This is what I tried, for the left-hand limit:

Since $c$ is a maximum we have $f(c) \geq f(c+h)$ for all $h$. Then:

$$\lim_{h \to 0-}\frac{f(c+h) - f(c)}{h} \geq 0 \iff \\\forall \epsilon>0, \exists \delta>0 : \forall h, \forall k \geq 0, 0 < 0-h < \delta \implies \left|\frac{f(c+h) - f(c)}{h}-k\right| < \epsilon$$

I don't even know if this is right but it was how I tried to structure the limit definition to accommodate a left-handed limit being $\leq 0$. But even in this form I wasn't able to prove it.

How does one prove these inequalities?

• Can you show that if $g(h) \geq 0$ for all $h$, and $\lim_{h\to 0} g(h)$ exists, then $\lim_{h\to 0} g(h) \geq 0$? – Matthew Leingang Mar 20 '18 at 17:17

Let $L=\lim_{h\rightarrow 0^{-}}\dfrac{f(c+h)-f(c)}{h}$, we are to show that $L\geq 0$.
Given $\epsilon>0$, there is a $\delta>0$, for $-\delta<h<0$, then $\left|\dfrac{f(c+h)-f(c)}{h}-L\right|<\epsilon$, then $L-\dfrac{f(c+h)-f(c)}{h}>-\epsilon$, then $L>\dfrac{f(c+h)-f(c)}{h}-\epsilon$.
Since $f$ has maximum at $x=c$, then $f(c+h)\leq f(c)$, so $f(c+h)-f(c)\leq 0$. Since $h<0$, so $\dfrac{f(c+h)-f(c)}{h}\geq 0$, so $L>-\epsilon$. Since this is true for all $\epsilon>0$, we have $L\geq 0$. If not, just put $\epsilon=-L$, then we end up with $L>-(-L)=L$, a contradiction.